Q.If $a_1, a_2, a_3, \ldots, a_r$ are in G.P., then prove that the determinant $\begin{vmatrix} a_{r+1} & a_{r+5} & a_{r+9} \\ a_{r+7} & a_{r+11} & a_{r+15} \\ a_{r+11} & a_{r+17} & a_{r+21} \end{vmatrix}$ is independent of $r$.

Showing the 12 most recent of 32 on this concept.

1. AP EAPCET 2021Set eng-2021-08-19-FN1 markMCQ
   Q.If $a_1, a_2, \ldots a_9$ are in G.P, then $\begin{vmatrix} \log a_1 & \log a_2 & \log a_3 \\ \log a_4 & \log a_5 & \log a_6 \\ \log a_7 & \log a_8 & \log a_9 \end{vmatrix}$ is equal to (A) $\log(a_1, a_2, \ldots a_n)$ (B) $1$ (C) $(\log a_9)^9$ (D) $0$

   ›Reveal solutionSolution

   The logs of GP terms form an arithmetic sequence, which makes the three rows of the determinant linearly dependent ($R_1+R_3=2R_2$); hence the determinant is $0$.

   Concept and Intuition

   Taking logs of a GP converts multiplication into addition, turning the GP into an AP. When a $3\times3$ matrix's rows are built from consecutive terms of one long AP (each row a block of 3 consecutive AP terms), the rows themselves form an AP of row-vectors — meaning the middle row is the average of the outer two. Any such linear dependency among rows forces determinant $=0$.

   Step-by-Step Solution

   1. Since $a_1,\dots,a_9$ are in GP with common ratio $r$: $a_n = a_1 r^{n-1}$, so $\log a_n = \log a_1 + (n-1)\log r$.
   2. Let $L=\log a_1$, $d=\log r$. Then $\log a_n = L+(n-1)d$.
   3. Row 1: $(\log a_1,\log a_2,\log a_3) = (L, L+d, L+2d)$. Row 2: $(\log a_4,\log a_5,\log a_6) = (L+3d, L+4d, L+5d)$. Row 3: $(\log a_7,\log a_8,\log a_9) = (L+6d, L+7d, L+8d)$. …

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2. AP EAPCET 2025Set eng-2025-05-23-FN1 markMCQ
   Q.$\begin{vmatrix} 2 & 1 \\ 3 & 1 \end{vmatrix} + \begin{vmatrix} 1 & 1/3 \\ 3 & 1 \end{vmatrix} + \begin{vmatrix} 1/2 & 1/9 \\ 3 & 1 \end{vmatrix} + \begin{vmatrix} 1/4 & 1/27 \\ 3 & 1 \end{vmatrix} + \ldots \infty =$ (A) $0$ (B) $\dfrac{1}{2}$ (C) $-\dfrac{1}{2}$ (D) $-1$

   ›Reveal solutionSolution

   Expand each $2\times2$ determinant, recognise two independent geometric series in the result, and sum them separately. Answer: (C).

   Concept and Intuition

   Every term in the sum is a $2\times2$ determinant with a fixed bottom row $\begin{pmatrix}3&1\end{pmatrix}$, so $\begin{vmatrix}a&b\\3&1\end{vmatrix}=a-3b$ splits linearly into an $a$-part and a $b$-part. Once each part is recognised as its own geometric progression, the infinite sum is just the difference of two standard geometric series sums, $\dfrac{\text{first term}}{1-r}$.

   Step-by-Step Solution

   1. General term: $\begin{vmatrix}a_n & b_n\\3&1\end{vmatrix} = a_n\cdot1 - b_n\cdot3 = a_n - 3b_n$.
   2. First-column values across the given terms: $2, 1, \tfrac12, \tfrac14,\dots$ — a geometric sequence with first term $2$ and ratio $\tfrac12$: $a_n = 2\left(\tfrac12\right)^{n-1}$.
   3. Second-column values: $1, \tfrac13, \tfrac19, \tfrac1{27},\dots$ — geometric with first term $1$ and ratio $\tfrac13$: $b_n=\left(\tfrac13\right)^{n-1}$. …

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3. AP EAPCET 2022Set eng-2022-07-04-AN1 markMCQ
   Q.If a, b, c are respectively the $5^{th}$, $8^{th}$, $13^{th}$ terms of an arithmetic progression, then $\begin{vmatrix} a & 5 & 1 \\ b & 8 & 1 \\ c & 13 & 1 \end{vmatrix} =$ (A) $0$ (B) $1$ (C) $abc$ (D) $520$

   ›Reveal solutionSolution

   Since $a,b,c$ are AP terms whose positions (5, 8, 13) are also linear in the same way, the determinant with columns (term, position, 1) is identically zero.

   Concept and Intuition

   If $a,b,c$ are terms of an AP at positions $p,q,r$ respectively, then $a=A+(p-1)d,\ b=A+(q-1)d,\ c=A+(r-1)d$ — i.e. each term is an affine (linear) function of its position. A determinant of the form $\begin{vmatrix}\text{term} & \text{position} & 1\end{vmatrix}$ over three such rows is always zero, because the "term" column is a linear combination of the "position" column and the "1" column (row reduction makes a column of zeros).

   Step-by-Step Solution

   1. Let the AP have first term $A$ and common difference $d$. Then $a = A+4d$ (5th term), $b=A+7d$ (8th term), $c=A+12d$ (13th term).
   2. Expand the determinant along the first row: $\Delta = a(8\cdot1-1\cdot13) - 5(b\cdot1-1\cdot c) + 1(b\cdot13-8\cdot c)$ $= a(-5) -5(b-c) + (13b-8c) = -5a -5b+5c+13b-8c = -5a+8b-3c$.
   3. Substitute: $-5a = -5A-20d$, $8b=8A+56d$, $-3c=-3A-36d$. …

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4. AP EAPCET 2021Set eng-2021-08-25-FN1 markMCQ
   Q.If $a, b, c$ are real numbers such that $a^2+b^2+c^2-ab-bc-ac \le 0$, then $\begin{vmatrix} (a-b+1)^5 & b^7-c^7 & c^9-a^9 \\ a^{11}-b^{11} & (b-c+2)^3 & c^{13}-a^{13} \\ a^{15}-b^{15} & b^{17}-c^{17} & (c-a+3)^1 \end{vmatrix} =$ (A) $2abc$ (B) $0$ (C) $24abc$ (D) $24$

   ›Reveal solutionSolution

   The inequality forces $a=b=c$; substituting collapses the matrix to a simple upper-triangular-looking form whose determinant is just the product of its surviving diagonal terms, $24$.

   Concept and Intuition

   The expression $a^2+b^2+c^2-ab-bc-ca$ is a sum of squares in disguise: $\tfrac12[(a-b)^2+(b-c)^2+(c-a)^2]$, which can never be negative. So the given condition "$\le 0$" combined with this built-in "$\ge0$" forces it to be exactly zero, which only happens when $a=b=c$. That single deduction massively simplifies every entry of the determinant, since almost every entry is a difference of equal powers of $a,b,c$ (which all vanish when $a=b=c$), leaving only three surviving diagonal-type terms.

   Step-by-Step Solution

   1. Rewrite the condition: $a^2+b^2+c^2-ab-bc-ca=\tfrac12[(a-b)^2+(b-c)^2+(c-a)^2]$, which is always $\ge0$.
   2. Given this quantity is also $\le0$, it must equal exactly $0$, forcing $(a-b)^2=(b-c)^2=(c-a)^2=0$, i.e. $a=b=c$.
   3. Substitute $a=b=c$ into the matrix: all terms of the form $b^7-c^7$, $c^9-a^9$, $a^{11}-b^{11}$, $c^{13}-a^{13}$, $a^{15}-b^{15}$, $b^{17}-c^{17}$ become $0$ (difference of equal quantities). …

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5. AP EAPCET 2021Set eng-2021-08-23-AN1 markMCQ
   Q.The value of the determinant $\begin{vmatrix} a+b & a+2b & a+3b \\ a+2b & a+3b & a+4b \\ a+4b & a+5b & a+6b \end{vmatrix}$ is ____ (A) $a$ (B) $b$ (C) $0$ (D) $a+b$

   ›Reveal solutionSolution

   The rows of this determinant are in arithmetic progression (each row's entries increase by a constant step, and consecutive rows shift by a constant amount too); row-reducing shows two rows become proportional, forcing the determinant to $0$.

   Concept and Intuition

   A determinant is zero whenever any two rows (or columns) are linearly dependent (e.g. one is a scalar multiple of another, or a linear combination of others). Rows built from an arithmetic-progression pattern (like $a+b, a+2b, a+3b$ then shifting by a constant each row) are a classic setup for this — subtracting consecutive rows collapses the "arithmetic" structure into constant, proportional rows.

   Step-by-Step Solution

   1. Original matrix:

   $\begin{vmatrix} a+b & a+2b & a+3b \\ a+2b & a+3b & a+4b \\ a+4b & a+5b & a+6b \end{vmatrix}$

   2. Perform $R_2 \to R_2 - R_1$: new $R_2 = (a+2b-(a+b),\ a+3b-(a+2b),\ a+4b-(a+3b)) = (b,\ b,\ b)$.
   3. Perform $R_3 \to R_3 - R_2^{\text{(original)}}$: new $R_3 = (a+4b-(a+2b),\ a+5b-(a+3b),\ a+6b-(a+4b)) = (2b,\ 2b,\ 2b)$. …

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6. AP EAPCET 2024Set eng-2024-05-19-AN1 markMCQ
   Q.$\begin{vmatrix} 1 & 1 & 1 \\ a^2 & b^2 & c^2 \\ a^3 & b^3 & c^3 \end{vmatrix} =$ (A) $(a-b)(b-c)(c-a)(a+b+c)$ (B) $(a-b)(b-c)(c-a)$ (C) $(a-b)(b-c)(a-c)(ab+bc+ca)$ (D) $(a-b)(b-c)(c-a)(ab+bc+ca)$

   ›Reveal solutionSolution

   This determinant is a known generalised-Vandermonde identity that factors as $(a-b)(b-c)(c-a)(ab+bc+ca)$ — verified numerically, option (D).

   Concept and Intuition

   Determinants with rows $1,x,x^2$ (Vandermonde) factor neatly as $(a-b)(b-c)(c-a)$. When the exponent pattern is not consecutive (here $0,2,3$ instead of $0,1,2$), the determinant still factors into the Vandermonde piece times an extra symmetric-polynomial factor — here that extra factor turns out to be $e_2 = ab+bc+ca$ (the second elementary symmetric polynomial), since the exponent set $\{0,2,3\}$ is the base set $\{0,1,2\}$ shifted up by the partition $(0,1,1)$, whose associated Schur polynomial is exactly $e_2$.

   Rather than rely purely on this identity from memory, verifying with actual numbers is the safest exam technique.

   Step-by-Step Solution

   1. Expand the determinant along the first row (all entries 1): $D = (b^2c^3-b^3c^2) - (a^2c^3-a^3c^2) + (a^2b^3-a^3b^2)$ $= b^2c^2(c-b) + a^2c^2(a-c) + a^2b^2(b-a)$.
   2. Rather than fully factor symbolically, test with concrete numbers: let $a=1, b=2, c=3$.
      • Direct determinant: rows $(1,1,1)$, $(1,4,9)$, $(1,8,27)$. $D = 1(4\cdot27-9\cdot8) - 1(1\cdot27-9\cdot1) + 1(1\cdot8-4\cdot1) = 1(108-72)-1(27-9)+1(8-4) = 36-18+4=22$.
   3. Test option (D): $(a-b)(b-c)(c-a)(ab+bc+ca)$ with these values: $(1-2)(2-3)(3-1) = (-1)(-1)(2) = 2$; $ab+bc+ca = 2+6+3=11$; product $=2\times11=22$. Matches $D=22$. …

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7. AP EAPCET 2025Set eng-2025-05-23-AN1 markMCQ
   Q.If $\alpha$ is a real root of the equation $x^3+6x^2+5x-42=0$, then the determinant of the matrix $\begin{bmatrix}\alpha-1 & \alpha+1 & \alpha+2\\ \alpha-2 & \alpha+3 & \alpha-3\\ \alpha+4 & \alpha-4 & \alpha+5\end{bmatrix}$ is (A) 90 (B) 120 (C) -105 (D) -135

   ›Reveal solutionSolution

   The cubic has a unique real root $\alpha=2$ (found by testing small integers and confirming the remaining quadratic has no real roots); substituting into the determinant and expanding gives $-105$.

   Concept and Intuition

   When a cubic is asked to have "a real root," it's often meant to be found by the rational root theorem (testing small divisors of the constant term) — here the constant is $-42$, and $2$ divides it. Once found, factor it out and check the resulting quadratic's discriminant to confirm no other real roots exist, so $\alpha$ is unambiguous.

   Step-by-Step Solution

   1. Test $x=2$ in $x^3+6x^2+5x-42$: $8+24+10-42=0$. So $x=2$ is a root.
   2. Factor: $x^3+6x^2+5x-42=(x-2)(x^2+8x+21)$.
   3. Discriminant of $x^2+8x+21$ is $64-84=-20<0$, so this factor has no real roots — confirming $\alpha=2$ is the unique real root.
   4. Substitute $\alpha=2$ into the matrix:

   $\begin{bmatrix}1&3&4\\0&5&-1\\6&-2&7\end{bmatrix}$

   5. Expand along row 1: …

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8. AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
   Q.For a fixed positive integer $n$, if $D = \begin{vmatrix} n! & (n+1)! & (n+2)! \\ (n+1)! & (n+2)! & (n+3)! \\ (n+2)! & (n+3)! & (n+4)! \end{vmatrix}$, then $\dfrac{D}{n!\,(n+1)!\,(n+2)!} =$ (A) $-4$ (B) $-2$ (C) $2$ (D) $4$

   ›Reveal solutionSolution

   Taking the factorials common from each row collapses the determinant to a constant: $\dfrac{D}{n!\,(n+1)!\,(n+2)!} = 2$ — option (C).

   Working. Factor $n!$, $(n+1)!$, $(n+2)!$ from rows 1, 2, 3 respectively:

   $D = n!\,(n+1)!\,(n+2)!\begin{vmatrix} 1 & n+1 & (n+1)(n+2) \\1 & n+2 & (n+2)(n+3) \\1 & n+3 & (n+3)(n+4) \end{vmatrix}$

   Hence $\dfrac{D}{n!\,(n+1)!\,(n+2)!}$ equals that $3\times3$ determinant. Apply $R_2\to R_2-R_1$ and $R_3\to R_3-R_1$, using $(n+2)(n+3)-(n+1)(n+2)=2(n+2)$ and $(n+3)(n+4)-(n+1)(n+2)=4n+10$: …

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9. AP EAPCET 2025Set eng-2025-05-26-FN1 markMCQ
   Q.If the inverse of $\begin{bmatrix}-x & 14x & 7x\\ 0 & 1 & 0\\ x & -4x & -2x\end{bmatrix}$ is $\begin{bmatrix}2 & 0 & 7\\ 0 & 1 & 0\\ 1 & -2 & 1\end{bmatrix}$, then $\begin{vmatrix}x & x+1 & x+2\\ x+1 & x+2 & x+3\\ x+2 & x+3 & x+4\end{vmatrix} =$ (A) $\dfrac{x}{5}$ (B) $x-5$ (C) $5x-1$ (D) $x+5$

   ›Reveal solutionSolution

   This tests recognizing a determinant with AP-rows (always zero) and pinning down $x$ from a matrix-inverse condition; the answer is whichever option numerically equals $0$ at that $x$.

   Concept and Intuition

   A square matrix whose rows form an arithmetic progression (i.e. $R_1-2R_2+R_3=0$) is always singular: its rows are linearly dependent, so its determinant is identically zero, no matter what parameter sits inside it. Spotting this saves a full cofactor expansion. The matrix-inverse condition is just the defining relation $A\,A^{-1}=I$ applied to one convenient entry.

   Step-by-Step Solution

   1. Let $A=\begin{bmatrix}-x & 14x & 7x\\ 0 & 1 & 0\\ x & -4x & -2x\end{bmatrix}$ and $A^{-1}=\begin{bmatrix}2 & 0 & 7\\ 0 & 1 & 0\\ 1 & -2 & 1\end{bmatrix}$.
   2. Using $A\,A^{-1}=I$, multiply row 1 of $A$ by column 1 of $A^{-1}$: $(-x)(2)+(14x)(0)+(7x)(1)=5x$. This must equal the $(1,1)$ entry of $I$, i.e. $1$. So $5x=1\Rightarrow x=\dfrac15$. (Checking the other products confirms this value is consistent throughout the matrix.)
   3. Now look at $D=\begin{vmatrix}x & x+1 & x+2\\ x+1 & x+2 & x+3\\ x+2 & x+3 & x+4\end{vmatrix}$. Apply $R_1\to R_1-2R_2+R_3$: each entry becomes $x-2(x+1)+(x+2)=0$, turning the first row into $(0,0,0)$.
   4. A zero row forces $D=0$ for every value of $x$ — this is an algebraic identity, independent of the specific $x=1/5$. …

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10. AP EAPCET 2024Set eng-2024-05-21-AN1 markMCQ
    Q.$\begin{vmatrix} a+b+2c & a & b \\ c & b+c+2a & b \\ c & a & c+a+2b \end{vmatrix} =$ (A) $(a+b+c)^3$ (B) $2(a+b+c)^3$ (C) $3(a+b+c)^3$ (D) $(a+b+c)$

    ›Reveal solutionSolution

    Writing each diagonal entry as $s+(\text{off-diagonal-like term})$ where $s=a+b+c$ reveals the matrix as $sI$ plus a rank-1 correction, whose determinant works out cleanly to $2(a+b+c)^3$ using the determinant lemma (or row operations).

    Concept and Intuition

    Many "cyclic-symmetric" determinants like this one simplify beautifully once you notice $a+b+2c = (a+b+c)+c$, etc. — subtracting off the common sum $s=a+b+c$ from every diagonal entry turns the matrix into $sI$ (a multiple of the identity) plus a very simple rank-1 matrix whose every row is identical. Determinants of $sI + \text{rank-1}$ matrices have a clean closed form.

    Step-by-Step Solution

    1. Let $s=a+b+c$. Rewrite diagonal entries: $a+b+2c=s+c$, $b+c+2a=s+a$, $c+a+2b=s+b$.
    2. The matrix becomes:

    $M=\begin{bmatrix}s+c & a & b\\ c & s+a & b\\ c & a & s+b\end{bmatrix}$

    3. Subtract $sI$: $M - sI = \begin{bmatrix}c&a&b\\c&a&b\\c&a&b\end{bmatrix}$ — every row is the same vector $(c,a,b)$, so $M-sI$ has rank 1: $M-sI=\mathbf 1 v^T$ where $\mathbf 1=(1,1,1)^T$ and $v=(c,a,b)^T$.
    4. So $M=sI+\mathbf1v^T$. The determinant lemma (matrix determinant lemma) gives: $\det(sI+\mathbf1 v^T) = s^3 + s^2\,(v^T\mathbf1)$ …

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11. AP EAPCET 2021Set eng-2021-08-20-AN1 markMCQ
    Q.If $k \in R$ and $\det A = \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix} = k$ then $\det B = \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 + 2a_1 & b_2 + 2b_1 & c_2 + 2c_1 \\ a_3 & b_3 & c_3 \end{vmatrix}$ is equal to (A) $0$ (B) $2K$ (C) $K$ (D) $K^2$

    ›Reveal solutionSolution

    $\det B$ is obtained from $\det A$ by the elementary row operation $R_2 \to R_2 + 2R_1$, which

    leaves a determinant unchanged; so $\det B = \det A = k$.

    Concept and Intuition

    A core determinant property: adding a scalar multiple of one row to another row does not change the determinant's value. This is because splitting the determinant using linearity in

    row 2 gives the original determinant plus $2\times$ a determinant with two identical rows (rows 1

    and the added copy of row 1), and any determinant with two identical rows is exactly zero.

    Step-by-Step Solution

    1. $\det B = \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2+2a_1 & b_2+2b_1 & c_2+2c_1 \\ a_3 & b_3 & c_3 \end{vmatrix}$.
    2. By linearity in row 2, split this into two determinants: $\det B = \begin{vmatrix} a_1&b_1&c_1\\a_2&b_2&c_2\\a_3&b_3&c_3\end{vmatrix} + 2\begin{vmatrix} a_1&b_1&c_1\\a_1&b_1&c_1\\a_3&b_3&c_3\end{vmatrix}$. …

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12. AP EAPCET 2021Set eng-2021-08-23-FN1 markMCQ
    Q.If $\omega$ is a root of the equation $x + \dfrac{1}{x} + 1 = 0$, then $\begin{vmatrix} 1 & 1+\omega & 1+\omega+\omega^2 \\ 3 & 4+3\omega & 5+4\omega+3\omega^2 \\ 6 & 9+6\omega & 11+9\omega+6\omega^2 \end{vmatrix} =$ (A) $1$ (B) $-1$ (C) $0$ (D) $1+\omega$

    ›Reveal solutionSolution

    Using $1+\omega+\omega^2=0$ to simplify the matrix entries and row-reducing gives determinant $-1$.

    Concept and Intuition

    $\omega$ here is a complex cube root of unity ($\omega^2+\omega+1=0$, $\omega^3=1$). Many determinant entries are partial sums of $1,\omega,\omega^2$ multiples, so replacing $1+\omega+\omega^2$ by $0$ wherever it appears collapses the algebra dramatically.

    Step-by-Step Solution

    1. Entry $(1,3)=1+\omega+\omega^2=0$.
    2. Entry $(2,3)=5+4\omega+3\omega^2$. Since $3\omega^2=3(-1-\omega)=-3-3\omega$, this equals $5+4\omega-3-3\omega=2+\omega$.
    3. Entry $(3,3)=11+9\omega+6\omega^2=11+9\omega+6(-1-\omega)=5+3\omega$.
    4. The matrix is now $\begin{pmatrix}1 & 1+\omega & 0\\ 3 & 4+3\omega & 2+\omega\\ 6 & 9+6\omega & 5+3\omega\end{pmatrix}$.
    5. Row reduce: $R_2\to R_2-3R_1 = (0,\,1,\,2+\omega)$; $R_3\to R_3-6R_1=(0,\,3,\,5+3\omega)$. …

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